Rotational Kinetic Energy of merry-go-round

A 31.0 kg child runs with a speed of 2.80 m/s tangential to the rim of a stationary merry-go-round . The merry-go-round has a moment of inertia of 520 kg\cdot m^2 and a radius of 2.51 m. When the child jumps onto the merry-go-round, the entire system begins to rotate.

A) Calculate the initial kinetic energy of the system.

B) Calculate the final kinetic energy of the system.

2. Relevant equations
E=1/2*m*v^2+1/2*I*w^2

3. The attempt at a solution

I got the answer to part A by simply doing KE=1/2*m*v^2. The answer was 122 J.
I'm not sure how to approach part B, however. I thought that because of conservation of mechanical energy that the initial energy would equal the final and I could use my answer from part A to solve part B. But this isn't working. Any help would be appreciated. :D

The KE energy is not the same as it was initially. There must be some interaction between the child and merry-go-round to set it into rotation and this consumes some energy. You can use conservation of angular momentum to calculate angular velocity, and calculate the rotational energy from that.